Question: 49: Given below are half cell reactions:
\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \mathrm{E}_{\mathrm{Mn}^{2+} / \mathrm{MnO}_{4}^{-}}^{\circ}=-1.510 \mathrm{~V} \frac{1}{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O} \mathrm{E}_{\mathrm{O}_{2} / \mathrm{H}_{2} \mathrm{O}}^{\circ}=+1.223 \mathrm{~V}Will the permanganate ion, \mathrm{MnO}_{4}^{-} liberate \mathrm{O}_{2} from water in the presence of an acid?
(1) Yes, because \mathrm{E}_{\text {cell }}^{\circ}=+2.733 \mathrm{~V}
(2) No, because E_{\text {cell }}^{\circ}=-2.733 \mathrm{~V}
(3) Yes, because \mathrm{E}_{\text {cell }}^{\circ}=+0.287 \mathrm{~V}
(4) No, because E_{\text {cell }}^{\circ}=-0.287 \mathrm{~V}
Answer: Option (3)
Explanation:
To check whether the reaction is feasible, the standard cell potential
E_{\text{cell}}^{\circ} is calculated.
The given value \mathrm{E}_{\mathrm{Mn}^{2+} / \mathrm{MnO}_{4}^{-}}^{\circ}=-1.510 \mathrm{~V} represents the oxidation potential of the permanganate couple.
The reduction potential for the oxygen electrode is \mathrm{E}_{\mathrm{O}_{2} / \mathrm{H}_{2} \mathrm{O}}^{\circ}=+1.223 \mathrm{~V}.
The standard cell potential is given by:
\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {oxidation }}^{\circ}+\mathrm{E}_{\text {reduction }}^{\circ} \mathrm{E}_{\text {cell }}^{\circ}=(-1.510)+(+1.223) \mathrm{E}_{\text {cell }}^{\circ}=+0.287 \mathrm{~V}Since the value of \mathrm{E}_{\text {cell }}^{\circ} is positive, the reaction is spontaneous.
Therefore, permanganate ion can liberate oxygen from water in acidic medium.